Perimeter=1

Drawing altitude $\overline{DF}$ and few more lines, we form the symmetries about $\overline{AD}$ and $\overline{DB}$ in the diagram, which show that

• $|DE| = |DF| = |DC|$
• $\angle E = \angle AFD$ and $\angle BFD = \angle C$
• $|EA| = |AF|$ and $|FB| = |BC|$ , so that $\color{#D61F06}|EA| + |BC| = |AF| + |FB| = |AB|$
• So $\Delta DEA \cong \Delta AFD$ and $\Delta FBD \cong \Delta DBC$

Since the perimeter of the whole polygon is $4$ , then $\begin{array}{rlcccl} {\color{#D61F06}|AB|} + |CD| + |DE| + {\color{#D61F06}\left(|EA| + |AF|\right)} &= 4 & & & & {\color{#3D99F6}\text{Definition of perimeter.}}\\ {\color{#D61F06}|AB|} + {\color{#D61F06}|AB|} + {\color{#D61F06}|AB|} + {\color{#D61F06}|AB|} &= 4 & & & & {\color{#3D99F6}\text{Using the previous info}}\\ {\color{#D61F06}|AB|} &= 1 \end{array}$

Thus, since the pentagon can visually be presented as the square of side length $1$ (by flipping $\Delta DEA$ and $\Delta DBC$ ), the area of the given polygon is $\boxed{1}$ .